Chapter 1: Relativity
Special Relativity: “all motion is relative”
All motion is described relative to an inertial frame of reference
Inertial frame: frame of reference in which Newton’s law of
inertia holds
Inertial frames are (mathematically) simpler in Special Relativity, and are
related in a (mathematically simple) manner. Accelerations and forces are
accounted for in S.R.
General Relativity = Special Relativity + Gravity (“curved space time” =
more complex relations between inertial reference frames)
Postulates of Special Relativity
The laws of physics are the same in all inertial frames of
reference. (mathmatical form, conservation laws, etc.)
The speed of light in free space has the same value for all
inertial frames of reference. v ~ 3.00E8 m/s
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The First Postulate and “everyday” relativity:
The laws of physics are the same in all inertial frames of reference.
y
dx '
y’
vx' 
 vx - v
x’ = x - vt
dt '
dy '
y’ = y
v y' 
 vy
x
dt '
x’ z’ = z
dz '
v
z
vz' 
 vz
z’
t’ = t
dt '
Conservation of momentum, kinetic energy, etc
But waves like sound waves have “special” inertial frame,
the frame at rest w.r.t. medium
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Second Postulate: The speed of light in free space has the same
value for all inertial frames of reference
need a new relationship between coordinates of different inertial
frames
linear, invertible “transformation” =
Lorentz Transformations
x - vt
x ' vt '
x' 
x
2
2
1 - v c
1 - v c
y'  y
z'  z
t' 
y  y'
z  z'
t - v cx
1 - v c
2
t
t 'v cx '
1 - v c
2
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Time Dialation: the time interval between events depends upon the
inertial observer
proper time t0: the time interval between two events in a frame of
reference where the events occur at the same place.
timing with a moving mirror
c t/2
c t0/2
L0
L0
v
v t/2
v
c t 0 2  L0
c t 2
2
 L0  v t 2
2
2
 t
t0
1 - v c
2
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Time Dialation and the passage of time
t
t0
1 - v c
2
t0 = time interval for clock at rest relative to an observer =
proper time
t = time interval for clock in motion relative to an observer,
t > t0
v = speed of relative motion
c = speed of light
Example 1.1: A spacecraft is moving relative to the earth. An observer finds that,
according to her clock, 3601s elapse between 1 PM and 2 PM on the spacecrafts’s
clock. What is the spacecraft’s speed relative to the earth?
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Doppler Effect: frequency shifts due to moving source/observer
doppler effect for sound n  n  1  v / c 
0
 1 - V / c
n0 = emitted frequency
n = detected frequency
v = speed of observer
V = speed of source
c = speed of sound
But, light does not require a medium!
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Doppler effect with light: three cases
1- observer moving perpendicular to a line between him and light
source
t
source
time between ticks is dilated
t0
1 - v c
1
n
t
2
 n  n 0 1 - v c
2
1
n0 
t0
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Doppler effect with light: three cases
2- observer moving away from source
next tick travels farther
source
time between ticks is dilated
1 - v c 
n
t0
v

t
t 0 1  
2
1 - v c 
 c
vt  v 

1 - v c 1  v c 
T  t   t 1   
 n0
c
1  v c 2
 c
1
1
1- v c
n0  , n 
n  n0
t0
T
1 v c
2
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Doppler effect with light: three cases
2- observer moving away towards source
source
time between ticks is dilated
next tick travels less distance
1 - v c
n
 v
t 0 1 - 
 c
2
t
t0
1 - v c
2
vt
 v
T  t -  t 1 - 
 c
c
1
1
n 0  ,n 
t0
T

 n0
n  n0
1 - v c1  v c
1 - v c2
1 v c
1- v c
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Example 1.2: A driver is caught going through a red light. The driver claims that the
color she actually saw was green (n = 5.60x1014 Hz) and not red (n 0= 4.80x1014 Hz)
because of the doppler effect. How fast must she have been going?
Example 1.3: A distant galaxy Hydra is receding grom the earth at 6.12x107 m/s. By
how much is a green spectral line of wavelength l = 500 nm emitted by this galaxy
shifted towards the red end of the spectrum?
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Length Contraction: determining the extent of a moving object
v
v
L0 =vt
t
L =vt0
t0
1 - v c
 L  L0 1- v c
2
2
L  vt0  vt 1 - v c
2
Example: Muons created in earth’s atmosphere by cosmic rays have speeds of about
2.994x108 m/s(.998c) At rest, a muon lifetime is about 2.2 ms.
-How far would these muons travel in 2.2 ms?
-What is the time dialated lifetime? How far do they travel in THIS amount of time?
-What is the length contracted distance of the previous answer, as seen in the muons
frame of reference?
-How far does the earth move relative to the muons during the 2.2 mslifetime?
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The Twin “Paradox”: a counter-intuitive result
Two twins on earth are separated temporarily when one twin
takes a trip at .80c to a star twenty light years away.
Earth bound twin: round trip takes 2x(20 ly / .8ly/y) = 50 years
Traveling twin: Earth-Star distance is length contracted to
L  L0 1- v c  20ly 1- .8  12ly
2
2
so round trip takes 2x(12 ly / .8ly/y) = 30 years
Twins age differently! Travelling twin does not stay in one
inertial reference frame, situation is not symetric.
Example 1.4: Each twin emits a radio signal once a year for the duration of the voyage.
How many signals does each twin recieve?
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Electricity and Magnetism
Integral Form
  Qenc
 E  dA 
o
Maxwell’s Equations
Differential Form
  4
 E 
o
 
 B  0




dE
 
d E
  B  m o ( Jc   o
)
B

dl

m
(
I


)
o
c
o

dt
dt

 
dB
 
d M


E

 E  dl  - dt
dt
Lorentz Force Law F  q(E  v  B )
 
 B  dA 0
Mathematically invariant formulation
electric charge is a relativistic invariant quantity
2 Ey 1 2 Ey
Maxwell’s equations -> Wave Equation
 2
2
x
c t 2 cp351c1:13
Relativistic Inertia (“relativistic mass”)
consider elastic collision between two identical (when at rest) masses
y
V’B
y’
Y
VA
S
x
S’
z
z’
x’
Ball A moves vertically only in frame S with
speed VA , Ball B moves vertically only in
frame S’ with speed VB ’= VA . Ball A
rebounds in S with speed VA , Ball B rebounds
in S’ with speed VB’.
v
Y/2
Y/2
Collision in S
V A T0  2 Y 2
Collision in S’
VB 'T0  Y
momentum conservation: mA V A  mBVB
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momentum conservation in S :
mAV A  mBV B
VB  Y T
T  T0
1- v c
2
mA Y T0  mB Y  T0

1- v c 

2
mB  mA 1- v c
2
" Relativist ic Mass" (Relativistic Inertia)
m  m0
1 - v c 
2
m0  rest mass
Example 1.5: Find the mass of an electron (m0 = 9.1E-31 kg)whose speed is .99c.
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Relativistic Momentum
m0 v
p  mv 
2
1- v c
Newton's Second Law
dp dmv
dv
F

m
dt
dt
dt
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Relativistic Kinetic Energy
v
d mv
KE   Fds  
ds   d mvv
dt
0
0
0
s
s
v
 mv -  mv dv 
2


1- v c
0
 m0 c 2 1- v c
2
1- v c
2
m0 v 2
1- v c
KE 
-
2
0
m0 v 2
v
m0 v 2
2
m0 v
1- v c
2
dv
v
0
 m0 c
m0 c 2
1- v c
2
2
1- v c - m0 c 2
2
- m0 c 2  mc 2 - m0 c 2
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KE  mc 2 - m0 c 2
mc2  KE  m0 c 2  E  KE  E0
E  mc 2  total energy
m0 c 2
1- v c
2
E0  m0 c 2  rest energy
at low speeds(v  c)
KE 
m0 c 2
1- v c
2

2

- m0 c  m0 c  1- v c

2
2

-1 2

- 1

 1
 1
2
KE  m0 c  1 v c  -1  m0 v 2
  2
 2
2
Correspondence Principle!
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Example 1.6: A stationary body explodes into two fragments each of mass 1kg that
move apart at speeds 0.6c relative to the original body. Find the rest mass of the
original body.
Example 1.7: Solar energy reaches the earth at a rate of about 1.4kW/m2. Given that
the average radius of the earth’s orbit is 1.5E11 m, how much mass does the sun lose
per second?
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General momentum-energy relations
E
m0 c 2
p
1- v c
2
2
E2 - p 2 c2 
m0 c
m0 v
1- v c
2
2
4
1- v c
2
-
m0 v 2 c 2
1- v c
2
m0 c 4  v 2 
2 4


1

m
0 c
2
2
1- v c  c 
2
2
E  p c  m0 c 4
2
2 2
for a massless particle (photon,neutrino, ...)
E  pc
(and v  c always)
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Electronvolts: typical energy scale for atomic physics
1eV = 1.6E-19 Coulomb x 1 Volt = 1.6E-19 J
X-Ray energies ~ keV
Nuclear Physics ~ MeV
Particle Physics ~ GeV
Example : What are the momenta and speeds of a 5 MeV
-electron? (m0 = .511 MeV/c2,
-proton? (m0 = 938 MeV/c2,
-photon? (m0 = 0 MeV/c2).
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Chapter 1 problems: 1,2,4,5,7,9,10,11,12,16,19,20,22,32,33,34,35,36,41,47,49
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